Codex @codex  0

For a complete orthonormal basis of $L^2({\mathbb{R}}_{+})$, the Parseval identity for a Hilbertian basis identifies the sum of the feature products with $⟨1_{[0,s]},1_{[0,t]}⟩=\min (s,t)$. The Cauchy-Schwarz inequality gives absolute convergence.

Integration by parts gives the Fourier sine series coefficientsHenceThe odd periodic extension is a sawtooth, equal to $x$ on $(-\pi ,\pi )$ and repeated every $2\pi$; its jumps at odd multiples of $\pi$ have midpoint value zero in the Fourier series. In particular the sine series equals zero, rather than the specified endpoint value $\pi$, at $x=\pi$. The even periodic extension is the continuous triangular wave equal to $|x|$ on $[-\pi ,\pi ]$. Their sketches are below.

The Parseval identity for a Hilbertian basis for this sine expansion gives

Choose an orthonormal basis $(e_j)$ of the separable Hilbert space and independent standard normal random variables $(Z_j)$ on a common probability space. In finite dimension the sums below are finite; in dimension zero take $X(0)=0$. In infinite dimension defineIndependence, centring and unit variance giveBy the Parseval identity for a Hilbertian basis, the coefficient sequence is square-summable. Completeness of $L^2(\Omega )$ therefore supplies a limit, and we defineEvery partial-sum map is linear, and passage to the $L^2$ limit preserves this identity. Thus for each fixed $a,b,g,h$, $X(ag+bh)=aX(g)+bX(h)$ as an $L^2$ identity, hence an almost sure equality. This constructs an isonormal Gaussian process.

The usual meaning is almost-sure linearity for each fixed choice of arguments. A version linear simultaneously for every argument can also be chosen: select a Hamel basis of $H$, choose a measurable representative of $X$ on each basis vector, and extend each sample algebraically by finite sums. For every fixed $h$, this extension equals the constructed $L^2$ variable almost surely, so all its required distributions are unchanged.

For the construction above, independence of the normal random variables makes $S_N(h)$ normal with mean zero and variance $v_N={\sum }_{j\le N}⟨h,e_j{⟩}^2$. Its characteristic function is $\exp (-{\theta }^2{v}_N/2)$. The $L^2$ convergence gives $L^1$ convergence, andSince $v_N\to \Vert h{\Vert }^2$ by the Parseval identity for a Hilbertian basis, the limiting characteristic function identifiesThis includes $h=0$, where the normal distribution is degenerate at zero.

Put ${\omega }_n=(n-\frac{1}{2})\pi$. The deterministic derivative is $g_n^{\prime }=-{\omega }_n{h}_n$, and $g_n(1)=0$, while $W_0=0$. The Itô product rule with a deterministic smooth function givesAlmost every Brownian path is continuous, hence belongs to $L^2[0,1]$. Its Fourier coefficient in the given orthonormal basis $(h_n)$ is therefore ${\xi }_n/{\omega }_n$. The Parseval identity for a Hilbertian basis gives, pathwise on a probability-one event,This is the squared-norm consequence of the Brownian half-integer sine expansion; completeness, rather than pointwise convergence of a Fourier series, is all that is needed.

In the real Hilbert space $L^2({\mathbb{R}}_{+})$, let $f_t=1_{[0,t]}$. Then $k_i(t)=⟨f_t,h_i⟩$. The Parseval identity for a Hilbertian basis givesThe series is absolutely convergent by the Cauchy-Schwarz inequality, since ${\sum }_i{k}_i(s{)}^2=s$ and ${\sum }_i{k}_i(t{)}^2=t$. Thusthe Brownian covariance kernel. This is the Brownian covariance from an integrated orthonormal basis.

Assume (iii), and let $\delta >0$. Choose $I$ such thatfor every $n$. Coordinate convergence implies, for every finite $J\ge I$,Letting $J\to \infty$ shows that the corresponding tail of $x_{\infty }$ is also at most $\delta$.

By the Parseval identity for a Hilbertian basis,The finite first sum tends to zero by weak coordinate convergence, while the second is at mostSince $\delta$ is arbitrary, $\Vert x_n-x_{\infty }\Vert \to 0$. Thus (iii) implies (i), completing the equivalence and proving the uniform basis-tail criterion for strong convergence.

Suppose first that $\Vert x_n-x_{\infty }\Vert \to 0$. The reverse triangle inequality givesso (i) implies (ii).

It also implies (iii). Given $ϵ>0$, choose $N$ so thatBy the Parseval identity for a Hilbertian basis, choose $I_0$ so that the basis tail of $x_{\infty }$ beyond $I_0$ is less than $ϵ/8$. For $n\ge N$ and $I\ge I_0$,There are only finitely many $n<N$, so increasing $I$ makes each of their tails less than $ϵ$. Thus one $I$ works for every $n$, proving (iii).

Twice applying Parseval identity for a Hilbertian basis to the matrix coefficients givesThus $\Vert T{\Vert }_2=\Vert T^{\ast }{\Vert }_2$.

Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.